\subsection{异分母的分式加减法}\label{subsec:8-8}
\begin{enhancedline}

与异分母的分数加减法类似，\zhongdian{异分母的分式相加减，先通分，变为同分母的分式，然后再加减。}用式子表示是：
\begin{center}
    \setlength{\fboxsep}{.6em}
    \framebox{\quad
        $\dfrac{a}{b} \pm \dfrac{c}{d} = \dfrac{ad}{bd} \pm \dfrac{bc}{bd} = \dfrac{ad \pm bc}{bd} \juhao$
        \;}
\end{center}

\liti 计算 $\dfrac{5}{6a^2b} - \dfrac{2}{3ab^2} + \dfrac{3}{4abc}$。

\jie $\begin{aligned}[t]
        & \dfrac{5}{6a^2b} - \dfrac{2}{3ab^2} + \dfrac{3}{4abc} \\
    ={} & \dfrac{10bc}{12a^2b^2c} - \dfrac{8ac}{12a^2b^2c} + \dfrac{9ab}{12a^2b^2c} \\
    ={} & \dfrac{10bc - 8ac + 9ab}{12a^2b^2c} \juhao
\end{aligned}$

\liti 计算 $\dfrac{12}{m^2 - 9} + \dfrac{2}{3 - m}$。

\jie $\begin{aligned}[t]
    & \dfrac{12}{m^2 - 9} + \dfrac{2}{3 - m} = \dfrac{12}{(m + 3)(m - 3)} - \dfrac{2}{m - 3} \\
    & = \dfrac{12}{(m + 3)(m - 3)} - \dfrac{2(m + 3)}{(m + 3)(m - 3)} \\
    & = \dfrac{12 - 2(m + 3)}{(m + 3)(m - 3)} = \dfrac{12 - 2m - 6}{(m + 3)(m - 3)} \\
    & = \dfrac{-2m + 6}{(m + 3)(m - 3)} = \dfrac{-2(m - 3)}{(m + 3)(m - 3)} \\
    & = -\dfrac{2}{m + 3} \juhao
\end{aligned}$

\liti 计算 $a + 2 - \dfrac{4}{2 - a}$。

\jie $\begin{aligned}[t]
        & a + 2 - \dfrac{4}{2 - a} = \dfrac{a + 2}{1} + \dfrac{4}{a - 2} \\
    ={} & \dfrac{a^2 - 4}{a - 2} + \dfrac{4}{a - 2} = \dfrac{a^2}{a - 2} \juhao
\end{aligned}$

\liti 计算 $\left(\dfrac{x + 2}{x^2 - 2x} - \dfrac{x - 1}{x^2 - 4x + 4}\right) \div \dfrac{x - 4}{x}$。

\jie $\begin{aligned}[t]
        & \left(\dfrac{x + 2}{x^2 - 2x} - \dfrac{x - 1}{x^2 - 4x + 4}\right) \div \dfrac{x - 4}{x} \\
    ={} & \left[\dfrac{x + 2}{x(x - 2)} - \dfrac{x - 1}{(x - 2)^2}\right] \cdot \dfrac{x}{x - 4} \\
    ={} & \dfrac{(x + 2)(x - 2) - (x - 1)x}{x(x - 2)^2} \cdot \dfrac{x}{x - 4} \\
    ={} & \dfrac{x^2 - 4 - x^2 + x}{(x - 2)^2(x - 4)} = \dfrac{1}{(x - 2)^2} \juhao
\end{aligned}$

\liti 计算 $\dfrac{1}{x + 1} - \dfrac{x + 3}{x^2 - 1} \cdot \dfrac{x^2 - 2x + 1}{x^2 + 4x + 3}$。

\jie $\begin{aligned}[t]
        & \dfrac{1}{x + 1} - \dfrac{x + 3}{x^2 - 1} \cdot \dfrac{x^2 - 2x + 1}{x^2 + 4x + 3} \\
    ={} & \dfrac{1}{x + 1} - \dfrac{(x + 3)}{(x + 1)(x - 1)} \cdot \dfrac{(x - 1)^2}{(x + 1)(x + 3)} \\
    ={} & \dfrac{1}{x + 1} - \dfrac{x - 1}{(x + 1)^2} = \dfrac{x + 1 - x + 1}{(x + 1)^2} = \dfrac{2}{(x + 1)^2} \juhao
\end{aligned}$

\liti 汽车从甲地开往乙地，每小时行驶 $v_1$ 千米，$t$ 小时可以到达，
如果每小时多行驶 $v_2$ 千米，那么可以提前几小时到达？

\jie 甲乙两地之间的距离是 $v_1t$千米。每小时多行驶 $v_2$ 千米，则每小时行驶 $(v_1 + v_2)$ 千米，
从甲地到乙地需要 $\dfrac{v_1t}{v_1 + v_2}$ 小时。可以提前的小时数就是

$\begin{aligned}[t]
        & t - \dfrac{v_1t}{v_1 + v_2} = \dfrac{t}{1} - \dfrac{v_1t}{v_1 + v_2} = \dfrac{t(v_1 + v_2)}{v_1 + v_2} - \dfrac{v_1t}{v_1 + v_2} \\
    ={} & \dfrac{v_1t + v_2t - v_1t}{v_1 + v_2} = \dfrac{v_2t}{v_1 + v_2} \juhao
\end{aligned}$

答：可以提前 $\dfrac{v_2t}{v_1 + v_2}$ 小时到达。

\lianxi
\begin{xiaotis}

\xiaoti{计算：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}, rows={rowsep+=.25em}}
        \xxt{$\dfrac{1}{R_1} + \dfrac{1}{R_2}$；} & \xxt{$\dfrac{a}{b} - \dfrac{b}{a} - \dfrac{a^2 + b^2}{ab}$；} \\
        \xxt{$\dfrac{b^2}{4a^2} - \dfrac{c}{a}$；} & \xxt{$a - b + \dfrac{2b^2}{a + b}$；} \\
        \xxt{$\dfrac{5a}{6b^2c} - \dfrac{7b}{12ac^2} + \dfrac{11c}{8a^2b}$。}
    \end{tblr}

\end{xiaoxiaotis}


\xiaoti{计算：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}, rows={rowsep+=.25em}}
        \xxt{$\dfrac{1}{x + 1} - \dfrac{1}{x - 1}$；} & \xxt{$\dfrac{a}{a - b} + \dfrac{b^2}{a(b - a)}$；} \\
        \xxt{$\dfrac{1}{m^2 - m} + \dfrac{m - 5}{2m^2 - 2}$；} & \xxt{$\dfrac{y}{x + y} + \dfrac{xy}{y^2 - x^2}$；} \\
        \xxt{$\dfrac{a + 5}{5a - 20} + \dfrac{5}{a^2 - 9a + 20}$；} & \xxt{$\dfrac{1}{x + 3} - \dfrac{6}{x^2 - 9} - \dfrac{x - 1}{6 + 2x}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{计算：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}, rows={rowsep+=.25em}}
        \xxt{$\left(\dfrac{x}{x - 2} - \dfrac{x}{x + 2}\right) \div \dfrac{4x}{2 - x}$；} & \xxt{$\left(\dfrac{a}{a - b} + \dfrac{b}{b - a}\right) \cdot \dfrac{ab}{a - b}$；} \\
        \xxt{$\left(1 - \dfrac{1}{1 - x}\right)\left(\dfrac{1}{x^3} - 1\right)$；} & \xxt{$\dfrac{3 - m}{2m - 4} \div \left(m + 2 - \dfrac{5}{m - 2}\right)$；} \\
        \xxt{$1 - \dfrac{a - b}{a + 2b} \div \dfrac{a^2 - b^2}{a^2 + 4ab + 4b^2}$。}
    \end{tblr}

\end{xiaoxiaotis}

\end{xiaotis}

\end{enhancedline}

